Question

How do you integrate `(x^2+8)/(x^2-5x+6)` using partial fractions?

Answer 1

`int (x^2+8)/(x^2-5x+6) dx = x + 17 lnabs(x-3) -12ln abs(x-2)+C`

Explanation:

Factorize the denominator:

`x^2-5x+6 = (x-2)(x-3)`

Before performing partial fraction decomposition of the rational function, the numerator must have lower degree than the denominator, so split the function as:

`(x^2+8)/(x^2-5x+6) = (x^2-5x+6+5x+2)/(x^2-5x+6) = 1+(5x+2)/(x^2-5x+6)`

Now:

`(5x+2)/(x^2-5x+6) = A/(x-2)+B/(x-3)`

`(5x+2)/(x^2-5x+6) = (A(x-3)+B(x-2))/((x-2)(x-3))`

`5x+2 = Ax-3A+Bx-2B`

`5x+2 = (A+B)x- (3A+2B)`

`{(A+B=5),(3A+2B=-2):}`

`{(A=-12),(B=17):}`

So:

`(5x+2)/(x^2-5x+6) = 17/(x-3)-12/(x-2)`

`(x^2+8)/(x^2-5x+6) = 1 + 17/(x-3)-12/(x-2)`

`int (x^2+8)/(x^2-5x+6) dx = int dx + 17 int dx/(x-3) -12int dx/(x-2)`

`int (x^2+8)/(x^2-5x+6) dx = x + 17 lnabs(x-3) -12ln abs(x-2)+C`

Answer 2

`x+17 ln | x-3| -12 ln |x-2| +C`

Explanation:

Dividing by the denominator leads to

`{x^2+8}/{x^2−5x+6} = 1+{5x+2}/{x^2−5x+6}`

Since `x^2−5x+6 = (x-3)(x-2)` we can write

`{5x+2}/{x^2−5x+6} = A/{x-3}+B/{x-2}`

Multiplying both sides by `x^2−5x+6` leads to

`5x+2 = A(x-2) + B(x-3)`

Substituting `x=3` and `x=2` in turn leads to the result `A=17, B=-12`, so that

`{x^2+8}/{x^2−5x+6} = 1+17/{x-3}-12/{x-2}`

Integrating this leads to the quoted result quickly!